{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-fraction-01",
  "mko_id": "mko-fraction",
  "title": {
    "zh-Hant": "分數：判定等值、化為最簡正分母形式與分數加法",
    "en": "Fractions: deciding equal value, normalizing to lowest terms with a positive denominator, and adding fractions"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-fraction-01",
    "source": {
      "mko_id": "mko-fraction",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "aa89902ad405685be331872579647cec33e7ab6792273fe60ad8fefe6129f919",
      "english_sha256": "ea8b15d28832a77ca243cb2c04a33ac67244aca632c895666d9a33fa9b4982d4"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "mul_inverse_nonzero",
      "zero_product",
      "frac_rules",
      "mul_assoc",
      "mul_comm",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "(1) 判定 (−6)/(−8) 與 9/12 是否表示同一個有理數；(2) 把 (−6)/(−8) 化成最簡、正分母的表示，並說明數值不變；(3) 計算 3/4+5/6，並說明「分子加分子、分母加分母」得到的 8/10 不是和；(4) 說明 3/0 不是合法的分數表示。",
      "en": "(1) Decide whether (−6)/(−8) and 9/12 represent the same rational number; (2) bring (−6)/(−8) to lowest terms with a positive denominator and show that the value is unchanged; (3) compute 3/4+5/6 and show that 'numerator plus numerator over denominator plus denominator', giving 8/10, is not the sum; (4) explain why 3/0 is not a valid fraction representation."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "p、q、r、s 是整數，所有作為分母的數都非零（MKO 假設 a1）；q≠0 時 p/q 表示有理數 p·(1/q)。",
          "en": "p, q, r and s are integers and every number used as a denominator is nonzero (MKO assumption a1); for q≠0, p/q denotes the rational number p·(1/q)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "要處理的表示：(−6)/(−8)、9/12、3/4、5/6、6/8，以及寫法 3/0。",
          "en": "The representations to handle: (−6)/(−8), 9/12, 3/4, 5/6 and 6/8, and the expression 3/0."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "mko_fraction_def"
        ],
        "before_tex": "\\frac{-6}{-8},\\ \\frac{9}{12}",
        "after_tex": "\\text{valid representations, since}\\ -8\\neq 0,\\ 12\\neq 0",
        "relation": "definition",
        "rule": "mko_fraction_def",
        "conditions": [
          "-8\\neq 0",
          "12\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依 MKO，整數 p 與非零整數 q 寫成的 p/q 才是合法表示。兩個分母 −8 與 12 都不是 0，所以兩者都表示有理數。分子或分母是負數並不影響合法性。",
          "en": "By the MKO, p/q with an integer p and a nonzero integer q is a valid representation. The denominators −8 and 12 are both nonzero, so both expressions represent rational numbers. A negative numerator or denominator does not affect validity."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_fraction_def"
          ],
          "explanation": {
            "zh-Hant": "此步只核對合法性條件（分母非零），不推出數值關係。",
            "en": "This step only checks the validity condition (nonzero denominators) and derives no relation between values."
          }
        },
        "leaf_basis": "mko_fraction_def + numeric_evaluation"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "L_cross"
        ],
        "before_tex": "\\frac{-6}{-8}=\\frac{9}{12}",
        "after_tex": "(-6)\\cdot 12=9\\cdot(-8)",
        "relation": "equivalent",
        "rule": "L_cross",
        "conditions": [
          "-8\\neq 0",
          "12\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_cross（p=−6、q=−8、r=9、s=12），兩個表示等值當且僅當交叉乘積相等。引理需要 q≠0 與 s≠0，S01 已確認。",
          "en": "By Lemma L_cross with p=−6, q=−8, r=9, s=12, the two representations have equal value exactly when the cross products agree. The lemma needs q≠0 and s≠0, which S01 checked."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_cross",
            "S01"
          ],
          "explanation": {
            "zh-Hant": "引理兩個方向都在 S02.2、S02.3 證明。",
            "en": "Both directions of the lemma are proved in S02.2 and S02.3."
          }
        },
        "leaf_basis": "L_cross"
      },
      {
        "id": "S02.1",
        "uses": [
          "zero_product"
        ],
        "before_tex": "q\\neq 0,\\ s\\neq 0",
        "after_tex": "qs\\neq 0",
        "relation": "implies",
        "rule": "zero_product",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依零乘積性質，qs=0 ⇔ q=0 或 s=0。兩者都不成立，所以 qs≠0，因此 1/(qs) 存在。",
          "en": "By the zero-product property, qs=0 ⇔ q=0 or s=0. Neither holds, so qs≠0, and therefore 1/(qs) exists."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "zero_product"
          ],
          "explanation": {
            "zh-Hant": "需要兩個分母各自非零；只要一個是 0，qs 就是 0。",
            "en": "Both denominators must be nonzero; if either is 0 then qs is 0."
          }
        },
        "leaf_basis": "zero_product + de_morgan_logic"
      },
      {
        "id": "S02.2",
        "uses": [
          "G1",
          "eq_congruence",
          "mul_inverse_nonzero"
        ],
        "before_tex": "\\frac{p}{q}=\\frac{r}{s}",
        "after_tex": "ps=rq",
        "relation": "implies",
        "rule": "eq_congruence",
        "conditions": [
          "q\\neq 0",
          "s\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "兩邊同乘 qs：左邊 p·(1/q)·q·s=p·1·s=ps，右邊 r·(1/s)·q·s=r·q·(s·(1/s))=rq（結合律、交換律、q·(1/q)=1、s·(1/s)=1）。所以 ps=rq。",
          "en": "Multiply both sides by qs: the left side is p·(1/q)·q·s=p·1·s=ps, and the right side is r·(1/s)·q·s=r·q·(s·(1/s))=rq (associativity, commutativity, q·(1/q)=1 and s·(1/s)=1). So ps=rq."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "eq_congruence",
            "mul_inverse_nonzero",
            "mul_assoc",
            "mul_comm"
          ],
          "explanation": {
            "zh-Hant": "化簡需要 1/q、1/s 存在，即 q≠0、s≠0。這是 ⇒ 方向；⇐ 方向在 S02.3。",
            "en": "The simplification needs 1/q and 1/s to exist, that is q≠0 and s≠0. This is the ⇒ direction; ⇐ is in S02.3."
          }
        },
        "leaf_basis": "eq_congruence + mul_inverse_nonzero + mul_assoc + mul_comm + mul_one"
      },
      {
        "id": "S02.3",
        "uses": [
          "S02.1",
          "eq_congruence",
          "frac_rules"
        ],
        "before_tex": "ps=rq",
        "after_tex": "\\frac{p}{q}=\\frac{r}{s}",
        "relation": "implies",
        "rule": "eq_congruence",
        "conditions": [
          "qs\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依 S02.1，qs≠0，兩邊同乘 1/(qs)：左邊 (ps)/(qs)=p/q（frac_rules，消去非零的 s），右邊 (rq)/(qs)=(rq)/(sq)=r/s（交換律後消去非零的 q）。所以 p/q=r/s。S02.1 到 S02.3 合起來證得引理 L_cross。",
          "en": "By S02.1, qs≠0; multiply both sides by 1/(qs): the left side is (ps)/(qs)=p/q (frac_rules, cancelling the nonzero s), and the right side is (rq)/(qs)=(rq)/(sq)=r/s (commutativity, then cancelling the nonzero q). So p/q=r/s. S02.1 to S02.3 together prove Lemma L_cross."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02.1",
            "frac_rules",
            "eq_congruence"
          ],
          "explanation": {
            "zh-Hant": "乘以 1/(qs) 需要 qs≠0；若容許零分母，ps=rq 不能推回任何數值關係（見錯誤控制二）。",
            "en": "Multiplying by 1/(qs) needs qs≠0; if zero denominators were allowed, ps=rq would not lead back to any relation between values (see error control 2)."
          }
        },
        "leaf_basis": "eq_congruence + frac_rules + mul_comm"
      },
      {
        "id": "S03",
        "uses": [
          "S02"
        ],
        "before_tex": "(-6)\\cdot 12=9\\cdot(-8)",
        "after_tex": "-72=-72",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "精確計算兩個交叉乘積：(−6)·12=−72，9·(−8)=−72。兩邊相等，所以交叉乘積等式為真。",
          "en": "Compute both cross products exactly: (−6)·12=−72 and 9·(−8)=−72. The two sides are equal, so the cross-product equation is true."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確求值不改變等式的真假（N1、N2）。",
            "en": "Exact evaluation does not change whether the equation holds (N1, N2)."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S02",
          "S03"
        ],
        "before_tex": "(-6)\\cdot 12=9\\cdot(-8)\\ \\text{true}",
        "after_tex": "\\frac{-6}{-8}=\\frac{9}{12}",
        "relation": "implies",
        "rule": "modus_ponens",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S02 的等價加上 S03 的「交叉乘積相等」，得 (−6)/(−8)=9/12：兩個分母不同、符號也不同的表示，表示同一個有理數。",
          "en": "The equivalence of S02 together with the true cross-product equation of S03 gives (−6)/(−8)=9/12: two representations with different denominators and different signs name the same rational number."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S03",
            "modus_ponens"
          ],
          "explanation": {
            "zh-Hant": "結論來自交叉乘積，不是看分母是否相同。",
            "en": "The conclusion comes from the cross products, not from comparing denominators."
          }
        },
        "leaf_basis": "modus_ponens + iff_intro"
      },
      {
        "id": "S05",
        "uses": [
          "S01",
          "frac_rules"
        ],
        "before_tex": "\\frac{-6}{-8}",
        "after_tex": "\\frac{(-6)(-1)}{(-8)(-1)}=\\frac{6}{8}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "k=-1\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "正分母正規化：依 frac_rules（a/b=(ak)/(bk)，k≠0），分子分母同乘 k=−1，得 6/8，分母變成正數。核對：(−6)·8=−48=6·(−8)，依 L_cross 數值相同。",
          "en": "Positive-denominator normalization: by frac_rules (a/b=(ak)/(bk) for k≠0), multiply numerator and denominator by k=−1 to get 6/8, whose denominator is positive. Check: (−6)·8=−48=6·(−8), so by L_cross the value is the same."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "L_cross"
          ],
          "explanation": {
            "zh-Hant": "反向再乘一次 −1 即回到原表示；需要 k≠0。",
            "en": "Multiplying by −1 again returns the original representation; k≠0 is needed."
          }
        },
        "leaf_basis": "frac_rules + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "gcd_def",
          "frac_rules"
        ],
        "before_tex": "\\frac{6}{8}",
        "after_tex": "\\frac{3\\cdot 2}{4\\cdot 2}=\\frac{3}{4}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "k=2\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "約分：6 的正因數是 1、2、3、6，8 的正因數是 1、2、4、8，共同的最大者是 2，所以 gcd(6,8)=2。6=3·2、8=4·2，依 frac_rules（k=2≠0）得 6/8=3/4。",
          "en": "Reduction: the positive divisors of 6 are 1, 2, 3, 6 and those of 8 are 1, 2, 4, 8; the largest common one is 2, so gcd(6,8)=2. Since 6=3·2 and 8=4·2, frac_rules with k=2≠0 gives 6/8=3/4."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "gcd_def"
          ],
          "explanation": {
            "zh-Hant": "反向是分子分母同乘 2；兩向都只用 k=2≠0。",
            "en": "The backward move multiplies numerator and denominator by 2; both directions use only k=2≠0."
          }
        },
        "leaf_basis": "frac_rules + gcd_def + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S04",
          "gcd_def",
          "frac_rules"
        ],
        "before_tex": "\\frac{9}{12}",
        "after_tex": "\\frac{3\\cdot 3}{4\\cdot 3}=\\frac{3}{4}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "k=3\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "gcd(9,12)=3（9 的正因數 1、3、9；12 的正因數 1、2、3、4、6、12），9=3·3、12=4·3，所以 9/12=3/4。兩個原表示都化到同一個 3/4，與 S04 一致。",
          "en": "gcd(9,12)=3 (the positive divisors of 9 are 1, 3, 9 and those of 12 are 1, 2, 3, 4, 6, 12), and 9=3·3, 12=4·3, so 9/12=3/4. Both original representations reduce to the same 3/4, in agreement with S04."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "gcd_def"
          ],
          "explanation": {
            "zh-Hant": "反向是同乘 3；k=3≠0。",
            "en": "The backward move multiplies by 3; k=3≠0."
          }
        },
        "leaf_basis": "frac_rules + gcd_def + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S06",
          "S07",
          "gcd_def"
        ],
        "before_tex": "\\frac{3}{4}",
        "after_tex": "4>0,\\ \\gcd(3,4)=1:\\ \\text{lowest terms, positive denominator}",
        "relation": "implies",
        "rule": "gcd_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "3 的正因數是 1 與 3，而 3 不整除 4，所以 gcd(3,4)=1；分母 4>0。因此 3/4 是最簡正分母形式。(−6)/(−8)、6/8、9/12、3/4 是同一個數值的四種表示：MKO 所說的約分與正規化只改變表示，不改變數值。",
          "en": "The positive divisors of 3 are 1 and 3, and 3 does not divide 4, so gcd(3,4)=1; the denominator 4 is positive. Hence 3/4 is in lowest terms with a positive denominator. (−6)/(−8), 6/8, 9/12 and 3/4 are four representations of one value: as the MKO says, reduction and normalization change the representation, not the value."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "gcd_def",
            "S06",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "此步只宣稱 3/4 滿足最簡正分母的條件；不宣稱一般有理數的最簡表示唯一（那需要另外的定理）。",
            "en": "This step claims only that 3/4 meets the lowest-terms conditions; it does not claim that lowest-terms representations of rationals are unique in general, which needs a further theorem."
          }
        },
        "leaf_basis": "gcd_def + divides_def + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "L_add"
        ],
        "before_tex": "\\frac{3}{4}+\\frac{5}{6}",
        "after_tex": "\\frac{3\\cdot 6+5\\cdot 4}{4\\cdot 6}=\\frac{38}{24}",
        "relation": "equivalent",
        "rule": "L_add",
        "conditions": [
          "4\\neq 0",
          "6\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_add（p=3、q=4、r=5、s=6）：分子 3·6+5·4=18+20=38，分母 4·6=24。",
          "en": "By Lemma L_add with p=3, q=4, r=5, s=6: numerator 3·6+5·4=18+20=38 and denominator 4·6=24."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_add"
          ],
          "explanation": {
            "zh-Hant": "引理是兩個數相等的恆等式，兩邊表示同一個有理數，雙向成立；引理在 S09.1、S09.2 證明。",
            "en": "The lemma is an equality of numbers, so both sides name the same rational and the step is two-way; the lemma is proved in S09.1 and S09.2."
          }
        },
        "leaf_basis": "L_add + numeric_evaluation"
      },
      {
        "id": "S09.1",
        "uses": [
          "frac_rules",
          "mul_comm"
        ],
        "before_tex": "\\frac{p}{q}+\\frac{r}{s}",
        "after_tex": "\\frac{ps}{qs}+\\frac{rq}{qs}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "q\\neq 0",
          "s\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依 frac_rules 的擴分（k≠0）：p/q=(ps)/(qs)（k=s≠0），r/s=(rq)/(sq)=(rq)/(qs)（k=q≠0，再用交換律 sq=qs）。兩個分數現在有相同的分母 qs。",
          "en": "By scaling in frac_rules (k≠0): p/q=(ps)/(qs) with k=s≠0, and r/s=(rq)/(sq)=(rq)/(qs) with k=q≠0 and commutativity sq=qs. The two fractions now share the denominator qs."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "mul_comm"
          ],
          "explanation": {
            "zh-Hant": "擴分與約分互為反運算，需要 s≠0 與 q≠0。",
            "en": "Scaling up and reducing undo each other; they need s≠0 and q≠0."
          }
        },
        "leaf_basis": "frac_rules + mul_comm"
      },
      {
        "id": "S09.2",
        "uses": [
          "S09.1",
          "S02.1",
          "frac_rules"
        ],
        "before_tex": "\\frac{ps}{qs}+\\frac{rq}{qs}",
        "after_tex": "\\frac{ps+rq}{qs}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "qs\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "同分母相加：a/c+b/c=(a+b)/c，其中 c=qs≠0（S02.1）。S09.1 與 S09.2 合起來證得引理 L_add。",
          "en": "Adding over a common denominator: a/c+b/c=(a+b)/c with c=qs≠0 (S02.1). S09.1 and S09.2 together prove Lemma L_add."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "S02.1"
          ],
          "explanation": {
            "zh-Hant": "這是對所有非零 c 成立的恆等式，雙向成立。",
            "en": "This is an identity for every nonzero c, so it holds both ways."
          }
        },
        "leaf_basis": "frac_rules + distributive"
      },
      {
        "id": "S10",
        "uses": [
          "S09",
          "gcd_def",
          "frac_rules"
        ],
        "before_tex": "\\frac{38}{24}",
        "after_tex": "\\frac{19\\cdot 2}{12\\cdot 2}=\\frac{19}{12}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "k=2\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "gcd(38,24)=2：38=2·19，19 是質數且不整除 24，所以共同正因數只有 1 與 2。38=19·2、24=12·2，約分得 19/12；gcd(19,12)=1，分母 12>0，這是最簡正分母形式。所以 3/4+5/6=19/12。",
          "en": "gcd(38,24)=2: 38=2·19 with 19 prime and not dividing 24, so the only common positive divisors are 1 and 2. Since 38=19·2 and 24=12·2, reduction gives 19/12; gcd(19,12)=1 and 12>0, so this is in lowest terms with a positive denominator. Hence 3/4+5/6=19/12."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "gcd_def"
          ],
          "explanation": {
            "zh-Hant": "約分改變表示不改變數值；N11 到 N13 重算。",
            "en": "Reduction changes the representation, not the value; N11 to N13 recompute it."
          }
        },
        "leaf_basis": "frac_rules + gcd_def + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S10",
          "L_cross"
        ],
        "before_tex": "\\frac{3+5}{4+6}=\\frac{8}{10}\\ \\text{vs}\\ \\frac{19}{12}",
        "after_tex": "8\\cdot 12=96\\neq 190=19\\cdot 10,\\ \\text{so}\\ \\frac{8}{10}\\neq\\frac{19}{12}",
        "relation": "implies",
        "rule": "L_cross",
        "conditions": [
          "10\\neq 0",
          "12\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "「分子加分子、分母加分母」得 8/10。依 L_cross，8/10=19/12 當且僅當 8·12=19·10；但 96≠190，所以 8/10≠19/12。這個文字操作不是分數加法（MKO 第三段）。",
          "en": "'Numerator plus numerator over denominator plus denominator' gives 8/10. By L_cross, 8/10=19/12 exactly when 8·12=19·10; but 96≠190, so 8/10≠19/12. This textual operation is not fraction addition (third paragraph of the MKO)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_cross",
            "S10",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "用的是 L_cross 的逆否：交叉乘積不等推出數值不等；兩個分母 10、12 都非零。",
            "en": "This uses L_cross in contrapositive form: unequal cross products give unequal values; both denominators 10 and 12 are nonzero."
          }
        },
        "leaf_basis": "L_cross + contrapositive + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "S05",
          "S06",
          "L_cross",
          "L_add"
        ],
        "before_tex": "\\frac{6}{8}\\ (=\\tfrac{3}{4})\\ \\text{and}\\ \\frac{5}{6}",
        "after_tex": "\\frac{6+5}{8+6}=\\frac{11}{14}\\neq\\frac{8}{10},\\quad \\frac{6\\cdot 6+5\\cdot 8}{8\\cdot 6}=\\frac{76}{48}=\\frac{19}{12}",
        "relation": "implies",
        "rule": "L_cross",
        "conditions": [
          "8\\neq 0",
          "6\\neq 0",
          "14\\neq 0",
          "48\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "把 3/4 換成等值的 6/8 再做文字操作，得 11/14；而 11·10=110≠112=8·14，所以 11/14≠8/10。同一對數值給出兩個不同結果，說明文字操作根本不是數的運算，只依賴寫法。真正的加法 L_add 給 76/48，且 76·12=912=19·48，仍等於 19/12：結果只依賴數值。",
          "en": "Replacing 3/4 by the equal 6/8 and repeating the textual operation gives 11/14; but 11·10=110≠112=8·14, so 11/14≠8/10. The same pair of values yields two different results, so the textual operation is not an operation on numbers at all; it depends on the notation. Genuine addition by L_add gives 76/48, and 76·12=912=19·48, so it is still 19/12: the result depends only on the values."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_cross",
            "L_add",
            "S06"
          ],
          "explanation": {
            "zh-Hant": "所有比較都經 L_cross，且所用分母 8、6、10、14、48、12 都非零。",
            "en": "Every comparison goes through L_cross, and all denominators used (8, 6, 10, 14, 48, 12) are nonzero."
          }
        },
        "leaf_basis": "L_cross + L_add + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "G1",
          "mul_inverse_nonzero",
          "mko_fraction_def"
        ],
        "before_tex": "\\frac{3}{0}",
        "after_tex": "\\text{not a valid representation: } 0\\ \\text{has no inverse}",
        "relation": "implies",
        "rule": "mul_inverse_nonzero",
        "conditions": [],
        "explanation": {
          "zh-Hant": "3/0 應表示 3·(1/0)，但 0 沒有倒數（對任何 t，0·t=0≠1），所以 3/0 不表示任何有理數，也不是「零」。MKO 只對 q≠0 定義 p/q。L_cross 也只對非零分母成立：若硬套到 3/0 與 6/0，交叉乘積 3·0=0=6·0 相等，卻沒有任何數值可比。",
          "en": "3/0 would mean 3·(1/0), but 0 has no inverse (0·t=0≠1 for every t), so 3/0 names no rational number, and in particular not zero. The MKO defines p/q only for q≠0. L_cross also holds only for nonzero denominators: forced onto 3/0 and 6/0, the cross products 3·0=0=6·0 agree, yet there are no values to compare."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mul_inverse_nonzero",
            "mko_fraction_def"
          ],
          "explanation": {
            "zh-Hant": "結論只說 3/0 不是合法表示；不宣稱交叉乘積可用來替它賦值。",
            "en": "The conclusion says only that 3/0 is not a valid representation; cross products are not claimed to give it a value."
          }
        },
        "leaf_basis": "mul_inverse_nonzero + mul_zero + mko_fraction_def"
      },
      {
        "id": "S14",
        "uses": [
          "S04",
          "S08",
          "S10",
          "S11",
          "S13"
        ],
        "before_tex": "S04,\\ S08,\\ S10,\\ S11,\\ S13",
        "after_tex": "\\frac{-6}{-8}=\\frac{9}{12}=\\frac{3}{4},\\ \\ \\frac{3}{4}+\\frac{5}{6}=\\frac{19}{12}\\neq\\frac{8}{10},\\ \\ \\frac{3}{0}\\ \\text{invalid}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "四個目標分別由 S04（等值）、S05 到 S08（正規化，數值不變）、S09 到 S11（加法與錯誤的文字操作）與 S13（零分母）完成。",
          "en": "The four goals are completed by S04 (equal value), S05 to S08 (normalization with unchanged value), S09 to S11 (addition and the wrong textual operation) and S13 (zero denominator)."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S04",
            "S08",
            "S10",
            "S11",
            "S13",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：(1) 到 (4)。足夠條件 H：S04、S08、S10、S11、S13 各自成立；H⇒G 由合取引入。H 由引理與精確計算正向證成，沒有把 G 當作前提。",
            "en": "Goal G: (1) to (4). Sufficient condition H: S04, S08, S10, S11 and S13 each hold; H⇒G by conjunction introduction. H is proved forwards from the lemmas and exact computation, and G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "(−6)/(−8)=9/12，因為 (−6)·12=−72=9·(−8)；它的最簡正分母形式是 3/4，約分與正規化只改變表示。3/4+5/6=(18+20)/24=38/24=19/12，而「分子分母分別相加」的 8/10 不等於 19/12，且結果隨寫法改變。3/0 不是合法表示，零分母不表示零。",
      "en": "(−6)/(−8)=9/12 because (−6)·12=−72=9·(−8); its lowest-terms positive-denominator form is 3/4, and reduction and normalization change only the representation. 3/4+5/6=(18+20)/24=38/24=19/12, while the 'add numerators and denominators separately' result 8/10 is not 19/12 and even changes with the notation. 3/0 is not a valid representation, and a zero denominator does not mean zero."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "9/12 與 (−6)/(−8) 的分母不同，所以數值不同",
          "en": "9/12 and (−6)/(−8) have different denominators, so their values differ"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "(−6)·12=−72=9·(−8)，依 L_cross 兩者相等。",
          "en": "(−6)·12=−72=9·(−8), so they are equal by L_cross."
        }
      },
      {
        "claim": {
          "zh-Hant": "3/0=0",
          "en": "3/0=0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "0 沒有倒數，3/0 不是合法表示；零值的表示是像 0/5 這樣分子為零、分母非零的分數。",
          "en": "0 has no inverse, so 3/0 is not a valid representation; zero is represented by fractions such as 0/5, with zero numerator and nonzero denominator."
        }
      },
      {
        "claim": {
          "zh-Hant": "3/4+5/6=(3+5)/(4+6)=4/5",
          "en": "3/4+5/6=(3+5)/(4+6)=4/5"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "和是 19/12；4·12=48≠95=19·5。",
          "en": "The sum is 19/12, and 4·12=48≠95=19·5."
        }
      },
      {
        "claim": {
          "zh-Hant": "因為 3·0=6·0，所以 3/0=6/0",
          "en": "Since 3·0=6·0, we have 3/0=6/0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "L_cross 需要分母非零；3/0、6/0 不表示任何數。",
          "en": "L_cross needs nonzero denominators; 3/0 and 6/0 name no numbers."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "zero_product",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "零乘積性質：在 ℤ、ℚ、ℝ、ℂ（或任何整域）中，uv=0 ⇔ u=0 或 v=0。",
          "en": "Zero-product property: in ℤ, ℚ, ℝ, ℂ (or any integral domain), uv=0 ⇔ u=0 or v=0."
        }
      },
      {
        "id": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "id": "mul_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法結合律：(uv)w=u(vw)。",
          "en": "Associativity of multiplication: (uv)w=u(vw)."
        }
      },
      {
        "id": "mul_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。",
          "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "de_morgan_logic",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "邏輯的德摩根律：¬(P∧Q) ⇔ ¬P∨¬Q；¬(P∨Q) ⇔ ¬P∧¬Q。",
          "en": "De Morgan's laws for logic: ¬(P∧Q) ⇔ ¬P∨¬Q and ¬(P∨Q) ⇔ ¬P∧¬Q."
        }
      },
      {
        "id": "eq_congruence",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等式同用：若 u=v，對兩邊做同一個有定義的運算 f，得 f(u)=f(v)。只對一邊做運算不屬於此規則。",
          "en": "Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule."
        }
      },
      {
        "id": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "id": "modus_ponens",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "肯定前件：由 P 與 P⇒Q 得 Q。",
          "en": "Modus ponens: from P and P⇒Q infer Q."
        }
      },
      {
        "id": "iff_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "雙條件：P⇒Q 與 Q⇒P 都成立時，P⇔Q 成立；反之亦然。",
          "en": "Biconditional: P⇔Q holds exactly when both P⇒Q and Q⇒P hold."
        }
      },
      {
        "id": "divides_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整除：對整數 a、b，a∣b 表示存在整數 k 使 b=ak。",
          "en": "Divisibility: for integers a, b, a∣b means there is an integer k with b=ak."
        }
      },
      {
        "id": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "id": "contrapositive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "逆否命題：P⇒Q 與 ¬Q⇒¬P 等價。逆命題 Q⇒P 一般不與 P⇒Q 等價。",
          "en": "Contrapositive: P⇒Q is equivalent to ¬Q⇒¬P. The converse Q⇒P is in general not equivalent to P⇒Q."
        }
      },
      {
        "id": "mul_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。",
          "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "mko_fraction_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-fraction",
          "sha256": "aa89902ad405685be331872579647cec33e7ab6792273fe60ad8fefe6129f919",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "對整數 p、q 且 q≠0，p/q 表示有理數。兩個合法表示 p/q 與 r/s 等值，當且僅當 ps=rq；約分與正分母正規化改變表示而不改變數值。",
          "en": "For integers p and q with q≠0, p/q represents a rational number. Two valid representations p/q and r/s have the same value exactly when ps=rq; reduction and positive-denominator normalization change the representation without changing the value."
        }
      },
      {
        "id": "gcd_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "對不全為 0 的整數 m、n，gcd(m,n) 是同時整除 m 與 n 的最大正整數；q>0 且 gcd(p,q)=1 時，稱 p/q 為最簡正分母形式。",
          "en": "For integers m, n not both 0, gcd(m,n) is the largest positive integer dividing both; p/q is in lowest terms with a positive denominator when q>0 and gcd(p,q)=1."
        }
      },
      {
        "id": "L_cross",
        "kind": "lemma",
        "proved_in": [
          "S02.1",
          "S02.2",
          "S02.3"
        ],
        "statement": {
          "zh-Hant": "對整數 p、q、r、s，q≠0、s≠0：p/q=r/s ⇔ ps=rq。",
          "en": "For integers p, q, r, s with q≠0 and s≠0: p/q=r/s ⇔ ps=rq."
        }
      },
      {
        "id": "L_add",
        "kind": "lemma",
        "proved_in": [
          "S09.1",
          "S09.2"
        ],
        "statement": {
          "zh-Hant": "對整數 p、q、r、s，q≠0、s≠0：p/q+r/s=(ps+rq)/(qs)。",
          "en": "For integers p, q, r, s with q≠0 and s≠0: p/q+r/s=(ps+rq)/(qs)."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "lhs": "(-6)*12",
        "rhs": "-72"
      },
      {
        "id": "N2",
        "step": "S03",
        "lhs": "9*(-8)",
        "rhs": "-72"
      },
      {
        "id": "N3",
        "step": "S05",
        "lhs": "(-6)*(-1)",
        "rhs": "6"
      },
      {
        "id": "N4",
        "step": "S05",
        "lhs": "(-8)*(-1)",
        "rhs": "8"
      },
      {
        "id": "N5",
        "step": "S05",
        "lhs": "(-6)*8",
        "rhs": "6*(-8)"
      },
      {
        "id": "N6",
        "step": "S06",
        "lhs": "gcd(6,8)",
        "rhs": "2"
      },
      {
        "id": "N7",
        "step": "S07",
        "lhs": "gcd(9,12)",
        "rhs": "3"
      },
      {
        "id": "N8",
        "step": "S08",
        "lhs": "gcd(3,4)",
        "rhs": "1"
      },
      {
        "id": "N9",
        "step": "S09",
        "lhs": "3*6+5*4",
        "rhs": "38"
      },
      {
        "id": "N10",
        "step": "S09",
        "lhs": "3/4+5/6",
        "rhs": "38/24"
      },
      {
        "id": "N11",
        "step": "S10",
        "lhs": "gcd(38,24)",
        "rhs": "2"
      },
      {
        "id": "N12",
        "step": "S10",
        "lhs": "38/24",
        "rhs": "19/12"
      },
      {
        "id": "N13",
        "step": "S10",
        "lhs": "gcd(19,12)",
        "rhs": "1"
      },
      {
        "id": "N14",
        "step": "S11",
        "lhs": "8*12",
        "relation": "!=",
        "rhs": "19*10"
      },
      {
        "id": "N15",
        "step": "S12",
        "lhs": "11*10",
        "relation": "!=",
        "rhs": "8*14"
      },
      {
        "id": "N16",
        "step": "S12",
        "lhs": "(6*6+5*8)*12",
        "rhs": "19*(8*6)"
      },
      {
        "id": "N17",
        "step": "S06",
        "vars": [
          "p",
          "q",
          "k"
        ],
        "degree": 2,
        "lhs": "p*(q*k)",
        "rhs": "(p*k)*q"
      },
      {
        "id": "N18",
        "step": "S13",
        "lhs": "3*0",
        "rhs": "6*0"
      },
      {
        "id": "N19",
        "lhs": "4*12",
        "relation": "!=",
        "rhs": "19*5"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內：分子分母為整數、分母非零（假設 a1），p/q 的數值是有理數 p·(1/q)。本稿由有理數的運算證明等值判準 ps=rq 與加法公式；不證明「每個有理數恰有一個最簡正分母表示」的一般唯一性（見開放項目）；不改動 canonical MKO 或英文 edition。",
      "en": "Inside the MKO's scope: integer numerators and denominators with nonzero denominators (assumption a1), the value of p/q being the rational number p·(1/q). The guide proves the equality criterion ps=rq and the addition formula from rational arithmetic; it does not prove the general uniqueness of the lowest-terms positive-denominator representation (see open items), and does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "\\frac{p}{q}",
        "分數表示，數值為 p·(1/q)",
        "a fraction representation, with value p·(1/q)"
      ],
      [
        "ps=rq",
        "交叉乘積等式",
        "the cross-product equation"
      ],
      [
        "\\gcd(m,n)",
        "m、n 的最大公因數",
        "the greatest common divisor of m and n"
      ],
      [
        "\\mathbb{Q}",
        "有理數",
        "the rational numbers"
      ]
    ],
    "md_error_controls": [
      {
        "title": "分母不同不代表數值不同 / Different denominators do not mean different values",
        "zh": "MKO 指出分數的字面表示不唯一。(−6)/(−8)、6/8、9/12、3/4 的分母各不相同，數值卻相同。判定兩個表示是否等值，要比較交叉乘積 ps 與 rq（L_cross），而不是比較分子或分母。",
        "en": "The MKO points out that the written form of a fraction is not unique. (−6)/(−8), 6/8, 9/12 and 3/4 all have different denominators but the same value. To decide whether two representations are equal, compare the cross products ps and rq (L_cross), not the numerators or denominators."
      },
      {
        "title": "零分母不表示零 / A zero denominator does not mean zero",
        "zh": "MKO 指出零分母不表示零。0/5 是零（分子為零、分母非零），但 3/0 根本不是表示，因為 0 沒有倒數。交叉乘積判準只在分母非零時成立：若對 3/0 與 6/0 硬套，3·0=6·0 會「證明」兩者相等，這是無意義的。",
        "en": "The MKO notes that a zero denominator does not mean zero. 0/5 is zero (zero numerator, nonzero denominator), but 3/0 is not a representation at all, because 0 has no inverse. The cross-product criterion holds only for nonzero denominators: forced onto 3/0 and 6/0, 3·0=6·0 would 'prove' them equal, which is meaningless."
      },
      {
        "title": "分子加分子、分母加分母不是分數加法 / Adding numerators and denominators separately is not fraction addition",
        "zh": "3/4 與 5/6 用文字操作得 8/10，但真正的和是 19/12，且 8·12=96≠190=19·10。更糟的是，把 3/4 寫成 6/8 再操作得 11/14≠8/10：結果依賴寫法，所以它不是數的運算。正確公式是 p/q+r/s=(ps+rq)/(qs)（L_add）。",
        "en": "The textual operation turns 3/4 and 5/6 into 8/10, but the true sum is 19/12, and 8·12=96≠190=19·10. Worse, writing 3/4 as 6/8 first gives 11/14≠8/10: the result depends on the notation, so it is not an operation on numbers. The correct formula is p/q+r/s=(ps+rq)/(qs) (L_add)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿只驗證 3/4 與 19/12 是最簡正分母形式，沒有證明一般的唯一性（每個有理數恰有一個最簡正分母表示）；那需要歐幾里得引理或貝祖等式，留給後續。",
        "en": "This guide only verifies that 3/4 and 19/12 are in lowest terms with positive denominators; it does not prove the general uniqueness (every rational has exactly one such representation), which needs Euclid's lemma or Bézout's identity and is left for later."
      },
      {
        "zh-Hant": "N17 是 p、q、k 的多項式恆等式核對，說明擴分保持交叉乘積相等；它補充 S05、S06，論證不依賴它。",
        "en": "N17 is a polynomial-identity check in p, q, k showing that scaling preserves cross-product equality; it supplements S05 and S06, and the argument does not depend on it."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "6d3a1d5f7536f8c32b52e78378a872546ab4eeaf24030466909289ff9ffaa771",
    "original_source_sha256": "6d3a1d5f7536f8c32b52e78378a872546ab4eeaf24030466909289ff9ffaa771",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}